Define $f(x, y, z) = yz^2$. Let $\vec{a} = (3, 1, -1)$ and $\vec{v} = \left( -1, 2, 0 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Answer: Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (3, 1, -1) + h \left( -1, 2, 0 \right) \right) - f(3, 1, -1)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left( 3 - h, 1 + 2h, -1 \right) - f(3, 1, -1)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{(1 + 2h)(-1)^2 - (1)(-1)^2}{h}$ We can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{1 + 2h - 1}{h}$ becomes $ \lim_{h \to 0} \dfrac{2h}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{2h}{h} &= \lim_{h \to 0} 2 \\ \\ &= 2 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = 2$.